$A$ bi-convex lens of focal length $10\,cm$ is cut into two identical parts along a plane perpendicular to the principal axis. The power of each lens after the cut is $...........\,D$.

Given a thin convex lens (refractive index $\mu_2$),kept in a liquid (refractive index $\mu_1, \mu_1 < \mu_2$) having radii of curvature $|R_1|$ and $|R_2|$. Its second surface is silver-polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?

An equiconvex lens of focal length $F$ is cut into two equal parts along the vertical axis. The focal length of each part will be

$A$ plano-convex lens $(f=20 \,cm)$ is silvered at its plane surface. What will be its new focal length (in $\,cm$)?

$A$ plano-convex lens,when silvered at its plane surface,is equivalent to a concave mirror of focal length $28 \, cm$. When its curved surface is silvered and the plane surface is not silvered,it is equivalent to a concave mirror of focal length $10 \, cm$. Find the refractive index of the material of the lens.

$A$ concave mirror of focal length $f_1$ is placed at a distance of $d$ from a convex lens of focal length $f_2$. $A$ beam of light coming from infinity and falling on this convex lens-concave mirror combination returns to infinity. The distance $d$ must equal: